Giải thích các bước giải:
a.Ta có : $ABCD$ là hình vuông
$\to BD\perp AC$
Mà $SA\perp (ABCD)\to SA\perp BD\to BD\perp (SAC)\to BO\perp (SAC)$
$\to \widehat{SB, SAC}=\widehat{BSO}$
Ta có : $AB=BC=CD=CA=SA=a\to BO=OD=OA=OC=\dfrac{a}{\sqrt{2}}$
$\to SO=\dfrac{a\sqrt{6}}{2}, SB=\sqrt{SA^2+AB^2}=a\sqrt{2}$
$\to \sin\widehat{OSB}=\dfrac{OB}{SB}=\dfrac{1}{2}\to\widehat{OSB}=30^o$
$\to\widehat{SB,SAC}=30^o$
b.Gọi $AE\perp SD=E$
Ta có : $CD\perp AD, SA\perp CD\to CD\perp (SAD)\to CD\perp AE$
Mà $AE\perp SD\to AE\perp (SCD)\to \widehat{CA,SCD}=\widehat{ACE}$
Ta có : $SA\perp AD, SA=AD=a\to AE=\dfrac{a\sqrt{2}}{2}$
$\to \sin\widehat{ACE}=\dfrac{AE}{AC}=\dfrac{1}{2}\to\widehat{ACE}=30^o$
$\to\widehat{CA,SCD}=30^o$
c.